JEE MAIN - Mathematics (2025 - 4th April Morning Shift - No. 23)
Let $A=\left[\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right]$. If for some $\theta \in(0, \pi), A^2=A^T$, then the sum of the diagonal elements of the matrix $(\mathrm{A}+\mathrm{I})^3+(\mathrm{A}-\mathrm{I})^3-6 \mathrm{~A}$ is equal to _________ .
Answer
6
Explanation
Note that $A$ is orthogonal:
$$A A^T=A^T A=I \text { and } A^T=A^{-1}$$
Given $A^2=A^T$, then:
$$\begin{aligned} & A^3=I \\ & \operatorname{Tr}(A+I)^3+(A-l)^3-6 A=\operatorname{Tr}\left(2 A^3+6 A-6 A\right) \\ & =\operatorname{Tr}\left(2 A^3\right)=\operatorname{Tr}(2 I) \\ & \left(\text { Using }(A+l)^3+(A-I)^3=2 A^3+6 A \text { and } 2 A^3=2 I\right)= \\ & 6 \end{aligned}$$
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